This is a proof from Apostol's Calculus. What I don't understand about it is the "Hence, there is at least one positive integer n such that n > b - 1" Is it because for every inductive set (the positive integers in this case) x+1 also belongs to the set? It doesn't specify why, and it doesn't seem like it. I guess it actually has something to do with some property of the least upper bound, but I don't understand how.
The relevant property here is that $b$ is the least upper bound.
Let us suppose on the contrary that there exists no $n \in \mathbb{Z}_+$ such that $n > b - 1$. In that case, we see that for all $n \in \mathbb{Z}_+$, $n \leq b - 1$. But this is exactly what it would mean for $b - 1$ to be an upper bound of $\mathbb{Z}_+$. $b - 1$ cannot be an upper bound of $\mathbb{Z}_+$, since $b$ is the least upper bound and $b - 1 < b$. This is a contradiction.
Therefore, there must be some $n \in \mathbb{Z}_+$ such that $n > b - 1$. And we therefore see that $n + 1 > b$, which contradicts the claim that $b$ is an upper bound.